For a summable function, with summable variation, prove that $\sup_{i \in I} \sup_{x \in [i]}|f(x)| \exp((t-1) sup_{x \in [i]}f(x) )$ is bounded

Question

Let $X = \mathbb{N}^\mathbb{N}$ and $f: X \to \mathbb{R}$ be a function such that

$$|f|_{var}= \sum_{n=1}^{\infty} var_n f < \infty,$$

where $var_n f = sup\{|f(x)-f(y)|: x,y \in X , x_k = y_k, \forall k = 1,\dots, n\}$. Also, suppose that $\sum_{i =1}^\infty \exp(\sup_{x \in [i]}f(x)) < \infty$, where $[i] = \{x \in X: x_1 = i\}$. We want to conclude that for $t>1$ exists a constant for $C_t$ such that

$$\sup_{i \in I} \sup_{x \in [i]}|f(x)| \exp( (t-1) sup_{x \in [i]}f(x)) < C_t.$$

To be more precise, I am trying to understand the proof of the Lemma 3.1 of the I. Morrison's following paper: "Entropy for zero-temperature limits of Gibbs-equilibrium states for countable-alphabet subshifts of finite type". But I think that understand this less general case, I understand the paper by myself. Thank you very much.

Answer 1

I have just found out that is very simple.

We have that $\sup f - \inf f \le \text{var}_1f$ and also that $\sup_{x \in [i]}f(x) \to - \infty $ as $ i \to \infty$.

Then, for $i$ sufficiently large \begin{align} \sup_{x \in [i]}|f(x)| \exp( (t-1) sup_{x \in [i]}f(x))& = \sup_{x \in [i]}(-f(x)) \exp( (t-1) sup_{x \in [i]}f(x))\\ &\le \left(\text{var}_1 f+ \inf_{x \in [i]}(-f(x)) \right) \exp( (t-1) sup_{x \in [i]}f(x))\\ &= \left(\text{var}_1 f + \sup_{x \in [i]}f(x) \right) \exp( (t-1) sup_{x \in [i]}f(x)). \end{align}

Hence, taking $i \to \infty$, the limit vanishes and the boundedness is guaranteed.